1 gram of hydrated oxalic acid `[(COOH)_(2).2H_(2)O]` undergoes combustion it produces 2.2 kcal of heat. What will be its enthalpy of combustion `(DeltaH_(c))`?

Options

A. `-2.2 kcal`
B. `-126kcal`
C. `-277.3kcal`
D. `-423kcal`

Answer

To find the enthalpy of combustion (ΔH_c) of hydrated oxalic acid \((COOH)_2 \cdot 2H_2O\), follow these steps:

### Step 1: Calculate the Molecular Weight of Hydrated Oxalic Acid
The molecular formula of hydrated oxalic acid is \((COOH)_2 \cdot 2H_2O\).

- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 4 from the oxalic acid + 4 from the water = 8 atoms × 1 g/mol = 8 g/mol
- Oxygen (O): 4 from the oxalic acid + 2 from the water = 6 atoms × 16 g/mol = 96 g/mol

Adding these together:
\[ \text{Molecular Weight} = 24 + 8 + 96 = 128 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Hydrated Oxalic Acid
Using the formula for moles:
\[ \text{Number of Moles} = \frac{\text{Weight}}{\text{Molecular Weight}} \]
Given that the weight is 1 gram:
\[ \text{Number of Moles} = \frac{1 \text{ g}}{128 \text{ g/mol}} = 0.0078125 \text{ moles} \]

### Step 3: Calculate the Enthalpy of Combustion
The enthalpy of combustion can be calculated using the formula:
\[ \Delta H_c = \frac{\text{Amount of Heat Released}}{\text{Number of Moles}} \]
Given that the amount of heat released is 2.2 kcal:
\[ \Delta H_c = \frac{2.2 \text{ kcal}}{0.0078125 \text{ moles}} \]
Calculating this gives:
\[ \Delta H_c = 281.6 \text{ kcal/mol} \]

### Step 4: Include the Sign for Exothermic Reaction
Since combustion is an exothermic reaction, we express the enthalpy change as negative:
\[ \Delta H_c = -281.6 \text{ kcal/mol} \]

### Final Answer
Thus, the enthalpy of combustion of hydrated oxalic acid is:
\[ \Delta H_c = -281.6 \text{ kcal/mol} \]